Applying Newton’s Laws Problem and Solutions 2

 Problem #5

Fig.5: Masses m1 and m2 are connected by a cord; m1 slides on frictionless slope.

A block of mass m₁ = 3.70 kg on a frictionless inclined plane of angle θ = 30.0⁰ is connected by a cord over a massless, frictionless pulley to a second block of mass m₂ = 2.30 kg hanging vertically, as shown in Figure. What are
(a) the magnitude of the acceleration of each block and
(b) the direction of the acceleration of m₂?
(c) What is the tension in the cord?

Answer:

Fig.6: The forces acting on m1

(a) Before thinking about the forces acting on these blocks, we can think about their motion. m1 is constrained to move along the slope and m₂ must move vertically. Because the two masses are joined by a string, the distance by which m1 moves up the slope is equal to the distance which m₂ moves downward, and the amount by which m1 moves down the slope is the amount by which m₂ moves upward. The same is true of their accelerations; if it turns out that m₁ is accelerating up the slope, that will be the same as m₂’s downward acceleration. Now we draw “free–body diagrams” and invoke Newton’s Second Law for each mass. Consider all the forces acting on m₁. These are shown in Figure.˙The force of gravity, with magnitude m1 pulls straight down. Here, looking ahead to the fact that motion can only occur along the slope it has decomposed into its components perpendicular to the surface (with magnitude m₁ cosθ) and down the slope (with magnitude m₁ sinθ). The normal force of the surface has magnitude N and points... normal to the surface! Finally the string pulls up with slope with a force of magnitude T, the tension in the string. Suppose we let x be a coordinate which measures movement up the slope. (Note, we are not saying that the block will move up the slope, this is just a choice of coordinate. Let y be a coordinate going perpendicular to the slope. We know that there is no y acceleration so the components of the forces in the y direction must add to zero. This gives

N − m₁g cos θ = 0 ⇒ N = m₁g cos θ

which gives the normal force should we ever need it. (We won’t.) Next, the sum of the x forces gives m1ax, which will not be zero. We get:

T − m₁g sin θ = m₁a_x (*)

Here there are two unknowns, T and a_x.

Fig.7:

The free–body diagram for mass m₂ is shown in Fig.7. The force of gravity, m₂g pulls downward and the string tension T pulls upward. Suppose we use a coordinate x₀ which points straight down. (This is a little unconventional, but you can see that there is a connection with the coordinate x used for the motion of m1. Then the sum of forces in the x’ direction gives m₂a_x’:

m₂g − T = m₂a_x’

Now as we argued above, the accelerations are equal: a_x = a_x’. This gives us:

m₂g − T = m₂a_x (**)

At this point, the physics is done and the rest of the problem is doing the math (algebra) to solve for ax and T. We are first interested in finding ax. We note that by adding Eqs. (*) and (**) we will eliminate T. Doing this, we get:

(T − m₁g sin θ) + (m₂g − T) = m₁a_x + m₂a_x

this gives:

m₂g − m₁g sin θ = (m₁ + m­₂)a_x

and finally:

a_x = (m₂ − m₁g sin θ)g/(m₁ + m₂)

Substituting the given values, we have:

a_x = (2.30 kg − 3.70 kg sin30⁰)(9.80 m/s²)/(3.70kg + 2.30kg)

    = +0 .735 m/s²

So a_x = +0 .735 m/s². What does this mean? It means that the acceleration of m1 up the slope and m2 downwards has magnitude 0.735 m/s². The plus sign in our result for ax is telling us that the acceleration does go in the way we (arbitrarily) set up the coordinates. If we had made the opposite (“wrong”) choice for the coordinates then our acceleration would have come out with a minus sign.

(b) We’ve already found the answer to this part in our understanding of the result for part (a). Mass m1 accelerates up the slope; mass m2 accelerates vertically downward.
(c) To get the tension in the string we may use either Eq. (*) or Eq. (**). Using (**) gives:
m₂g − T = m₂a_x ⇒ T = m₂g − m₂a_x = m₂(g − a_x)
Substituting everything,
T = (2 .30kg)(9.80 m/s² − (0.735 m/s²)) = 20.8 N\

Problem #6
A 10 kg monkey climbs up a massless rope that runs over a frictionless tree limb (!) and back down to a 15 kg package on the ground, as shown in Fig.8.

Fig.8: Monkey runs up the rope

(a) What is the magnitude of the least acceleration the monkey must have if it is to lift the package off the ground? If, after the package has been lifted the monkey stops its climb and holds onto the rope, what are
(b) the monkey’s acceleration and
(c) the tension in the rope?

Answer:
(a) Before we do anything else, let’s understand what forces are acting on the two masses in this problem. The free–body diagrams are shown in Fig. 9. The monkey holds onto the rope so it exerts an upward force of magnitude T, where T is the tension in the rope. Gravity pulls down on the monkey with a force of magnitude mg, where m is the mass of the monkey. These are all the forces. Note that they will not cancel since the problem talks about the monkey having an acceleration and so the net force on the monkey will not be zero. The forces acting on the box are also shown. Gravity pulls downward on the box with a force of magnitude Mg, M being the mass of the box. The rope pulls upward with a force T, If the box is resting on the ground, the ground will be pushing upward with some force F_ground. (Here, the ground cannot pull downward.) However when the box is not touching the ground then Fground will be zero.

Fig.9: The forces acting on the two masses

In the first part of the problem, the monkey is moving along the rope. It is not stuck to any point of the rope, so there is no obvious relation between the acceleration of the monkey and the acceleration of the box. Suppose we let a_y,monkey be the vertical acceleration of the monkey and ay,box be the vertical acceleration of the box. Then from our free–body diagrams, Newton’s Second Law gives the acceleration of the monkey:

T − mg = ma_y,monkey

When the box is on the ground its acceleration is zero and then T + F_gr = mg. But when the box is off then ground then we have:

T − Mg = Ma_box (Box off the ground)

In the first part of the problem we are solving for the condition that the monkey climbs just barely fast enough for the box to be lifted off the ground. If so, then the ground would exert no force but the net force on the box would be so small as to be virtually zero; the box has a very, very tiny acceleration upwards. From this we know:

T − Mg = 0 ⇒ T = Mg

and substituting this result into the first equation gives

Mg − mg = ma_monkey ⇒ a_monkey = (M − m)g/m

Substituting the given values,

a_monkey = (15kg − 10kg)(9.80 m/s²)/10kg = 4.9 m/s²

The monkey must pull himself upwards so as to give himself an acceleration of 4.9 m/s². Anything less and the box will remain on the ground. (b) Next, suppose that after climbing for while (during which time the box has been rising off the ground) the monkey grabs onto the rope. What new condition does this give us?
Now it is true that the distance that the monkey moves up is the same as the distance which the box moves down. The same is true of the velocities and accelerations of the monkey and box, so in this part of the problem (recalling that I defined both accelerations as being in the upward sense),

a_monkey = −a_box

This condition is not true in general, but here it is because we are told that the monkey is holding fast to the rope. If you recall the example of the Atwood machine from your textbook or lecture notes, this is the same physical situation we are dealing with here. We expect the less massive monkey to accelerate upwards and the more massive box to accelerate downwards. Let’s use the symbol a for the monkey’s vertical acceleration; then the box’s vertical acceleration is −a and our equations are:

T − mg = ma
and
T − Mg = M(−a)

At this point the physics is done and the rest is math (algebra) to solve for the two unknowns, T and a. Since the first of these equations gives T = mg + ma, substituting this into the second equation gives:

mg + ma − Mg = −Ma ⇒ ma + Ma = Mg – mg

which gives:
(M + m)a = (M − m)g
a = (M − m)/(M + m)g

Plugging in the numbers gives
When the monkey is holding tight to the rope and both masses move freely, the monkey’s acceleration is 2.0 m/s² upwards.
(c) Now that we have the acceleration a for this part of the problem, we can easily substitute into our results in part (b) and find the tension T. From T − mg = ma we get:

T = mg + ma = m(g + a) = (10.0 kg)(9.80 m/s² + 2.0 m/s²) = 118 N.

The tension in the rope is 118 N.

Problem #7
A mass M is held in place by an applied force F and a pulley system as shown in Fig. 10. The pulleys are massless and frictionless. Find
(a) the tension in each section of rope, T₁, T₂, T₃, T₄, and T₅, and
(b) the magnitude of F.

Fig.10

Answer:
(a) We note first that the mass M (and therefore everything else) is motionless. This simplifies the problem considerably! In particular, every mass in this problem has no acceleration and so the total force on each mass is zero. We have five rope tensions to find here, so we’d better start writing down some equations, fast! Actually, a few of them don’t take much work at all; we know that when we have the (idealized) situation of massless rope passing around a frictionless massless pulley, the string tension is the same on both sides. As shown in the figure, it is a single piece of rope that wraps around the big upper pulley and the lower one, so the tensions T₁, T₂ and T₃ must be the same:

T₁ = T₂ = T₃ Not bad so far!

Next, think about the forces acting on mass M. This is pretty simple... the force of gravity Mg pulls down, and the tension T₅ pulls upward. That’s all the forces but they sum to zero because M is motionless. So we must have

T₅ = Mg .

Next, consider the forces which act on the small pulley. These are diagrammed in Fig. 11(a). There is a downward pull of magnitude T₅ from the rope which is attached to M and also upward pulls of magnitude T₂ and T₃ from the long rope which is wrapped around the pulley. These forces must sum to zero, so T₂ + T₃ −T₅ =0 But we already know that T₅ = Mg and that T₂ = T₃ so this tells us that

2T₂ −Mg =0

which gives
T₂ = Mg/2 ⇒ T₃ = T₂ = Mg/2

Fig.11

We also have:
T₁ = T₂ = Mg/2.
Next, consider the forces on the large pulley, shown in Fig. 11(b). Tension T₄ (in the rope attached to the ceiling) pulls upward and tensions T1, T2 and T3 pull downward. These forces sum to zero, so

T₄ −T₁ −T₂ −T₃ = 0.

But T₄ is the only unknown in this equation. Using our previous answers,

T₄ = T₁ + T₂ + T₃ = Mg/2 + Mg/2 + Mg/2 = 3Mg/2

and so the answers are:
T₁ = T₂ = T₃ = Mg/2
T₄ =3Mg/2
T₅ = Mg

(b) The force F is the (downward) force of the hand on the rope. It has the same magnitude as the force of the rope on the hand, which is T₁, and we found this to be Mg/2. So F = Mg/2.

Problem #8
Mass m1 on a frictionless horizontal table is connected to mass m₂ through a massless pulley P₁ and a massless fixed pulley P₂ as shown in Fig. 12.
(a) If a₁ and a₂ are the magnitudes of the accelerations of m1 and m₂ respectively, what is the relationship between these accelerations? Find expressions for
(b) the tensions in the strings and
(c) the accelerations a₁ and a₂ in terms of m₁, m₂ and g.

Fig.12

Answer:
(a) Clearly, as m₂ falls, m1 will move to the right, pulled by the top string. But how do the magnitudes of the displacements, velocities and accelerations of m₂ and m1 compare? They are not necessarily the same. Indeed, they are not the same. Possibly the best way to show the relation between a₁ and a₂ is to do a little math; for a very complicated system we would have to do this anyway, and the practice won’t hurt.

Fig. 13

Focus on the upper mass (m₁) and pulley P1, and consider the lengths labelledin Fig. 4.14. x measures the distance from the wall to the moving pulley; clearly the position of m₂ is also measured by x. ` is the length of string from m₁ to the pulley. xblock measures the distance from the wall to m₁. Then:

x_block = x − l

This really ignores the bit of string that wraps around the pulley, but we can see that it won’t matter.
Now the total length of the string is L = x + l and it does not change with time. Since we have l = L− x, we can rewrite the last equation as

x_block = x − (L− x) = 2x − L

Take a couple time derivatives of this, keeping in mind that L is a constant. We get:

d²x_block/dt² = 2d²x/dt²

But the left side of this equation is the acceleration of m₁ and the right side is the (magnitude of the) acceleration of m₂. The acceleration of m1 is twice that of m₂:

a₁ = 2a₂

We can also understand this result by realizing that when m₂ moves down by a distance x, a length 2x of the string must go from the “underneath” section to the “above” section in Fig. 4.14. Mass m₁ follows the end of the string so it must move forward by a distance 2x. Its displacement is always twice that of m₂ so its acceleration is always twice that of m₂.

Fig.14:

(b) Now we try to get some information on the forces and accelerations, and we need to draw free–body diagrams. We do this in Fig. 4.15. Mass m₁ has forces m₁g acting downward, a normal force from the table N acting upward, and the string tension T₁ pulling to the right. The vertical forces cancel since m1 has only a horizontal acceleration, a₁. Newton’s Second Law gives us:

T₁ = m₁a₁   (1)

The pulley has forces acting on it, as shown in Fig. 14(b). The string wrapped around it exerts its pull (of magnitude T₁) both at the top and bottom so we have two forces of magnitude T₁ pulling to the left. The second string, which has a tension T₂, pulls to the right with a force of magnitude T₂. Now this is a slightly subtle point, but the forces on the pulley must add to zero because the pulley is assumed to be massless. (A net force on it would give it an infinite acceleration!) This condition gives us:

T₂ − 2T₁ = 0  (2)

Lastly, we come to m₂. It will accelerate downward with acceleration a₂. Summing the downward forces, Newton’s Second Law gives us:

m₂g −T₂ = m₂a₂   (3)

For good measure, we repeat the result found in part (a):

a₁ = 2a₂    (4)

In these equations, the unknowns are T₁, T₂, a₁ and a₂...four of them. And we have four equations relating them, namely Eqs. (1) through (4). The physics is done. We just do algebra to finish up the problem. There are many ways to do the algebra, but I’ll grind through it in following way: Substitute Eq. (4) into Eq. (1) and get:

T₁ = 2m₁a₂

Putting this result into Eq. (2) gives

T₂ − 2T₁ = T₂ − 4m₁a₂ = 0     

⇒ T₂ = 4m₁a₂ and finally using this in Eq. (3) gives

m₂g − 4m₁a₂ = m₂a₂ at which point we can solve for a₂ we find:

m₂g = a₂(4m₁ + m₂)

⇒ a₂ = m₂g/(4m₁ + m₂)            (5)

Having solved for one of the unknowns we can quickly find the rest. Eq. (4) gives us a1:

a₁ = 2a₂ = 2m₂g/(4m₁ + m₂)      (6)

Then Eq. (4) gives us T₁:

T₁ = m₁a₁ = 2m₁m₂g/(4m₁ + m₂)

Finally, since Eq. (2) tells us that T₂ = 2T₁ we get

T₂ = 4m₁m₂g/(4m₁ + m₂)          (7)

Summarizing our results from Eqs. (5) through (7), we have:

T₁ = 2m₁m₂g/(4m₁ + m₂)
T₂ = 4m₁m₂g/(4m₁ + m₂)

for the tensions in the two strings and:
(c) a₁ = 2m₂g/(4m₁ + m₂)
a₂ = m₂g/(4m₁ + m₂)

for the accelerations of the two masses. 

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